Path Integrals
Let's start off with a video. This is a great video which includes a complete introduction, with examples, and explanation of applications of path integrals. It is well worth your time to watch it.
video by Dr Chris Tisdell 

The equations from the last video are important for evaluating path integrals. Here is a summary.
requirements 
result  

continuous scalar function \(f(x,y,z)\) 
path integral of \(f\) over \( \mathcal{C} \)  
curve \( \mathcal{C} \) parameterized as \(\vec{c}(t)\) on an interval \(a \leq t \leq b\) 
\(\displaystyle{ \int_{\mathcal{C}}{ f(x,y,z)~ds} = }\) \(\displaystyle{ \int_{a}^{b}{ f(\vec{c}(t)) ~ \ \vec{c}'(t) \ ~dt } }\)  
continuous \(\vec{c}'(t)\) [derivative of \(\vec{c}(t)\) with respect to \(t\)] 
Most of the above equations should be familiar to you. However, one comment is in order about the term \(f(\vec{c}(t))\). How do you substitute a vector into a function?
If we write the vector as \( \vec{c}(t) = \langle X(t), Y(t), Z(t) \rangle = X(t) \hat{i} + Y(t) \hat{j} + Z(t)\hat{k} \) then we substitute the vector components into \(f(x,y,z)\) as \(x=X(t)\), \(y=Y(t)\) and \(z=Z(t)\). Another way to write this is
\(f(x,y,z) = f(X,Y,Z) = f(X(t), Y(t), Z(t))\). The result is a function of \(t\) only. All \(x\)'s, \(y\)'s and \(z\)'s will be gone, leaving only an integral in \(t\).
A very interesting comment he makes in the above video is that the direction of the curve does not affect the answer, i.e. integrating in the opposite direction will give the same answer. This is true because we use only the magnitude of the derivative of the parameterized curve, i.e. \(\ \vec{c}'(t) \\), which is the same regardless of the direction. However, the path integral is dependent on the chosen path, i.e. different paths will, in general, give different results. So path integrals are sensitive to the paths.
Okay, let's try some practice problems.
Practice
Unless otherwise instructed, evaluate these path integral over the curve \(\mathcal{C}\), giving your answers in exact terms.
Unless otherwise instructed, evaluate the path integral \( \int_{\mathcal{C}} x+y ~ds \) over the curve \( \mathcal{C}: \) straight line from \((0,0)\) to \((1,1)\), giving your answer in exact terms.
Problem Statement 

Unless otherwise instructed, evaluate the path integral \( \int_{\mathcal{C}} x+y ~ds \) over the curve \( \mathcal{C}: \) straight line from \((0,0)\) to \((1,1)\), giving your answer in exact terms.
Solution 

video by Dr Chris Tisdell 

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Unless otherwise instructed, evaluate the path integral \( \int_{\mathcal{C}}{x+y+z ~ ds } \) over the curve \( \mathcal{C}: \langle \cos(t), \sin(t), t \rangle; 0 \leq t \leq 2\pi \), giving your answer in exact terms.
Problem Statement 

Unless otherwise instructed, evaluate the path integral \( \int_{\mathcal{C}}{x+y+z ~ ds } \) over the curve \( \mathcal{C}: \langle \cos(t), \sin(t), t \rangle; 0 \leq t \leq 2\pi \), giving your answer in exact terms.
Solution 

video by Dr Chris Tisdell 

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Unless otherwise instructed, evaluate the path integral \( \int_{\mathcal{C}}{x+y+z ~ ds } \) over the curve \( \mathcal{C}: \) straight line \( (1,0,0) \) to \( ( 1,0, 2\pi) \), giving your answer in exact terms.
Problem Statement 

Unless otherwise instructed, evaluate the path integral \( \int_{\mathcal{C}}{x+y+z ~ ds } \) over the curve \( \mathcal{C}: \) straight line \( (1,0,0) \) to \( ( 1,0, 2\pi) \), giving your answer in exact terms.
Solution 

video by Dr Chris Tisdell 

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Unless otherwise instructed, evaluate the path integral \( f(x,y,z) = x^2+y^21+z \) over the curve \( \mathcal{C}: \langle \cos(t), \sin(t), t \rangle; 0 \leq t \leq 3\pi \), giving your answer in exact terms.
Problem Statement 

Unless otherwise instructed, evaluate the path integral \( f(x,y,z) = x^2+y^21+z \) over the curve \( \mathcal{C}: \langle \cos(t), \sin(t), t \rangle; 0 \leq t \leq 3\pi \), giving your answer in exact terms.
Solution 

video by Dr Chris Tisdell 

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Unless otherwise instructed, evaluate the path integral \( \int_{\mathcal{C}}{ x+y^2 ~ds }\) over the curve \( \mathcal{C}: \) 2 lines \((0,0)\)  \((1,1)\) and \((1,1)\)  \((1,0)\), giving your answer in exact terms.
Problem Statement 

Unless otherwise instructed, evaluate the path integral \( \int_{\mathcal{C}}{ x+y^2 ~ds }\) over the curve \( \mathcal{C}: \) 2 lines \((0,0)\)  \((1,1)\) and \((1,1)\)  \((1,0)\), giving your answer in exact terms.
Solution 

video by Dr Chris Tisdell 

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Unless otherwise instructed, evaluate the path integral \( \int_{\mathcal{C}}{xy^4 ~ds } \) over the curve \( \mathcal{C}: \) right half of the circle \( x^2+y^2 = 16 \), giving your answer in exact terms.
Problem Statement 

Unless otherwise instructed, evaluate the path integral \( \int_{\mathcal{C}}{xy^4 ~ds } \) over the curve \( \mathcal{C}: \) right half of the circle \( x^2+y^2 = 16 \), giving your answer in exact terms.
Solution 

video by PatrickJMT 

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Evaluate \( \int_{\mathcal{C}}{ 2x ~ds } \) where \(\mathcal{C} = \mathcal{C_1} + \mathcal{C_2}\); \(\mathcal{C_1}\) goes from \((0,0)\) to \((1,1)\) along \(y=x^2\); \(\mathcal{C_2}\) goes from \((1,1)\) to \((1,2)\) along \(x=1\).
Problem Statement 

Evaluate \( \int_{\mathcal{C}}{ 2x ~ds } \) where \(\mathcal{C} = \mathcal{C_1} + \mathcal{C_2}\); \(\mathcal{C_1}\) goes from \((0,0)\) to \((1,1)\) along \(y=x^2\); \(\mathcal{C_2}\) goes from \((1,1)\) to \((1,2)\) along \(x=1\).
Final Answer 

\((5\sqrt{5}+11)/6\)
Problem Statement 

Evaluate \( \int_{\mathcal{C}}{ 2x ~ds } \) where \(\mathcal{C} = \mathcal{C_1} + \mathcal{C_2}\); \(\mathcal{C_1}\) goes from \((0,0)\) to \((1,1)\) along \(y=x^2\); \(\mathcal{C_2}\) goes from \((1,1)\) to \((1,2)\) along \(x=1\).
Solution 

video by Michel vanBiezen 

Final Answer 

\((5\sqrt{5}+11)/6\) 
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Evaluate \(\int_{\mathcal{C}}{x^2z~ds}\), where \(\mathcal{C}\) is the line segment from \((0,6,2)\) to \((1,8,7)\).
Problem Statement 

Evaluate \(\int_{\mathcal{C}}{x^2z~ds}\), where \(\mathcal{C}\) is the line segment from \((0,6,2)\) to \((1,8,7)\).
Hint 

There are an infinite number of ways to parameterize this curve. This instructor uses the equations
\(x=(1t)x_1+tx_2\)
\(y=(1t)y_1+ty_2\)
\(z=(1t)z_1+tz_2\)
\(0 \leq t \leq 1\).
Problem Statement 

Evaluate \(\int_{\mathcal{C}}{x^2z~ds}\), where \(\mathcal{C}\) is the line segment from \((0,6,2)\) to \((1,8,7)\).
Hint 

There are an infinite number of ways to parameterize this curve. This instructor uses the equations
\(x=(1t)x_1+tx_2\)
\(y=(1t)y_1+ty_2\)
\(z=(1t)z_1+tz_2\)
\(0 \leq t \leq 1\).
Solution 

video by MIP4U 

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You CAN Ace Calculus
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The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
Set 1  basic identities  

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) 
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) 
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) 
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) 
Set 2  squared identities  

\( \sin^2t + \cos^2t = 1\) 
\( 1 + \tan^2t = \sec^2t\) 
\( 1 + \cot^2t = \csc^2t\) 
Set 3  doubleangle formulas  

\( \sin(2t) = 2\sin(t)\cos(t)\) 
\(\displaystyle{ \cos(2t) = \cos^2(t)  \sin^2(t) }\) 
Set 4  halfangle formulas  

\(\displaystyle{ \sin^2(t) = \frac{1\cos(2t)}{2} }\) 
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) 
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) 
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = \sin(t) }\)  
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) 
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = \csc^2(t) }\)  
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) 
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = \csc(t)\cot(t) }\) 
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\) 
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\)  
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) 
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = \frac{1}{1+t^2} }\)  
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
Trig Integrals
\(\int{\sin(x)~dx} = \cos(x)+C\) 
\(\int{\cos(x)~dx} = \sin(x)+C\)  
\(\int{\tan(x)~dx} = \ln\abs{\cos(x)}+C\) 
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)  
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) 
\(\int{\csc(x)~dx} = \) \( \ln\abs{\csc(x)+\cot(x)}+C\) 
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Practice Instructions
Unless otherwise instructed, evaluate these path integral over the curve \(\mathcal{C}\), giving your answers in exact terms.